qz5f08sol - October 10, 2008 PHY2054 Discussion-Fall 08...

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October 10, 2008 PHY2054 Discussion-Fall ‘08 Quiz 5 (Chapter 17.5-18.7) Name: UFID: *1. (5pts) Suppose that you wish to fabricate a uniform wire out of 2.00 g of copper. If the wire is to have a resistance of 0.300 Ω , and if all of the copper is to be used, what will be the length of the wire? The resistivity and density of copper are 1.70 × 10 -8 Ω m and 8.92 × 10 3 kg/m 3 . The volume of the copper is V = m/density = (2×10 -3 )/(8.92×10 3 ) = 2.24×10 -7 m 3 Since volume = cross sectional area × length, we can express resistance in terms of length and volume: R = ρL/A = ρL/(V/L) = ρL 2 /V L = √(RV/ρ) = √[0.3×2.24×10 -7 /(1.70×10 -8 )] = 1.99 m *2. (5pts) The heating coil of a hot water heater has a resistance of 8.00 Ω and it operates at 120 V. Assuming that the water absorbs all of the energy converted by the heating coil, calculate how long it takes to raise the temperature of 1.50 kg of water from 20.0 ˚ C to the boiling point? (Specific heat of water: C = 4.187
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qz5f08sol - October 10, 2008 PHY2054 Discussion-Fall 08...

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